In an $A.P.$ if $p^{th}$ term is $\large\frac{1}{q}$ and $q^{th}$ term is $\large\frac{1}{p}$, then prove that the sum of first $pq$ terms is $\large\frac{1}{2}$$(pq+1) where p\neq q 1 Answer Toolbox: • n^{th} term of an A.P.=t_n=a+(n-1)d where a=first term and d=common difference. • Sum of n terms of an A.P. is S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
Given:$p^{th}\:term=\large\frac{1}{q}$ and $q^{th}\:term=\large\frac{1}{p}$
$\Rightarrow\:t_p=\large\frac{1}{q}$ and $t_q=\large\frac{1}{p}$
$\Rightarrow\:a+(p-1)d=\large\frac{1}{q}$..........(i) and
$\Rightarrow\:a+(q-1)d=\large\frac{1}{p}$..........(ii)
Subtracting (i)-(ii) we get $d(p-q)=\large\frac{1}{q}-\frac{1}{p}=\frac{p-q}{pq}$
$\Rightarrow\:d=\large\frac{1}{pq}$
Substituting $d$ in (i) we get $a+\large\frac{p-1}{pq}=\frac{1}{q}$
$\Rightarrow\:a=\frac{1}{q}-\frac{p-1}{pq}=\frac{1}{pq}$
We know that $S_n=\large\frac{n}{2}$$[2a+(n-1)d] \therefore\:S_{pq}=\large\frac{pq}{2}$$[2a+(pq-1)d]$
$=\large\frac{pq}{2}$$[2\times \large\frac{1}{pq}$$+(pq-1)\large\frac{1}{pq}]$